Let $F=\langle a,b\rangle$ be a free group of rank $2$. Its commutator subgroup has a nice free-basis: $$[a^m,b^n], \,\,\,\,m,n\in\mathbb{Z}.$$

Instead of $[F,F]$, we consider another simplest normal subgroup. Let $N$ be the smallest normal subgroup of $F$ containing $a$. Of course, $N$ will not only contain powers of $a$ but elements $bab^{-1}$ also, and in general $waw^{-1}$ for any word $w$ in $F$.

Question: What is a free-basis of $N$?

I was thinking analogously that $\{b^iab^{-i} \colon i\in\mathbb{Z}\}$ would be a free basis. But, this is not, since in the product of these elements, the end points will be $b$ or $b^{-1}$; in particular, $(ab)a(ab)^{-1}$ can not be obtained from this set.

[As noticed by Guerin, in last paragraph, after "But, this ...." is incorrect!]

  • $\begingroup$ You have that $(ab)a(ab)^{-1}=b^0ab^{-0}bab^{-1}b^0a^{-1}b^{-0}$ so your element is in the set generated by your hypothetic base. $\endgroup$ – Clément Guérin Sep 24 '15 at 6:46
  • $\begingroup$ Oh! Thanks for pointing this. Then the mentioned set would be a generating set. Is it so? and is it "basis"? $\endgroup$ – Groups Sep 24 '15 at 6:48
  • $\begingroup$ Generating set property shouldn't be difficult, it suffices to show that for any $w$, $waw^{-1}$ can be written as a word in $b^iab^{-i}$ . I think this is easily handled by induction on the length of the word $w$. Free basis seems like a much harder point. $\endgroup$ – Clément Guérin Sep 24 '15 at 6:59
  • $\begingroup$ OK. I will try to prove at least that it is a generating set. $\endgroup$ – Groups Sep 24 '15 at 7:00

For the free basis property. One way to state this is that any word $w$ written as a reduced non-trivial word $w_0$ of $e_i:=b^iab^{-i}$'s is non trivial. Write your word $w_0(e_1,....)=e_{f(1)}^{\epsilon(1)}...e_{f(r)}^{\epsilon(r)}$.

I claim that if the length of $w_0(e_1,...)$ is (as a word in $a$'s and $b$'s) $l_0$ then provided that $f=b^ka^{\epsilon}b^{-k}$ is different from $e_{f(r)}^{-\epsilon(r)}$ then the length of $w_0(e_1,...)f$ is strictly greater than the length $l_0$.

Use this to show that if $length(w_0(e_1,...))=0$ (as a word in $a$'s and $b$'s) then $w_0(e_1,...)$ is a trivial word (up to reduction) in $e_i$'s.

What you have shown then is that the surjective (because of the generating set thing we discussed) morphism from $\mathbb{F}_{\mathbb{Z}}$ to $N$ sending $e_i$ to $b^iab^{-i}$ is one to one. Whence $\{b^iab^{-i}\}$ is a free basis.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.